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Fundamentals of physics 9e halliday resnick walker solutions manual and test bank

Fundamentals of physics 9e halliday resnick walker solutions manual and test bank

http://www.mediafire.com/view/gapb4o93ve8m64f/Fundamentals_of_Physics_9e_Halliday_Resnick_WalkerHalliday,_Renick_right_ch06.doc


http://www.mediafire.com/view/i5iu4isfn62314x/Fundamentals_of_Physics_9e_Halliday_Resnick_Walkerch02.rtf

Book Description

Publication Date: March 16, 2010 | ISBN-10: 0470469110 | ISBN-13: 978-0470469118 | Edition: 9
This book arms engineers with the tools to apply key physics concepts in the field. A number of the key figures in the new edition are revised to provide a more inviting and informative treatment. The figures are broken into component parts with supporting commentary so that they can more readily see the key ideas. Material from The Flying Circus is incorporated into the chapter opener puzzlers, sample problems, examples and end-of-chapter problems to make the subject more engaging. Checkpoints enable them to check their understanding of a question with some reasoning based on the narrative or sample problem they just read. Sample Problems also demonstrate how engineers can solve problems with reasoned solutions.
INCLUDES PARTS 1-4
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Import Settings:
Base Settings: Brownstone Default
Information Field: Difficulty
Information Field: Section
Highest Answer Letter: E
Multiple Keywords in Same Paragraph: No





Chapter: Chapter 2




Multiple Choice




1.  A particle moves along the x axis from xi to x f .  Of the following values of the initial and final coordinates, which results in the displacement with the largest magnitude?
A)  xi = 4m, x f = 6m
B)  xi = –4m, x f = –8m
C)  xi = –4m, x f = 2m
D)  xi = 4m, x f = –2m
E)  xi = –4m, x f = 4m

Ans:  E
Difficulty:  E
Section:  2-3




2.  A particle moves along the x axis from xi to x f .  Of the following values of the initial and final coordiantes, which results in a negative displacement?
A)  xi = 4m, x f = 6m
B)  xi = –4m, x f = –8m
C)  xi = –4m, x f = 2m
D)  xi = –4m, x f = –2m
E)  xi = –4m, x f = 4m

Ans:  B
Difficulty:  E
Section:  2-3




3.  The position y of a particle moving along the y axis depends on the time t according to the equation
y = atbt2. The dimensions of the quantities a and b are respectively:
A)  L2/T, L3/T2
B)  L/T2, L2/T
C)  L/T, L/T2
D)  L3/T, T2/L
E)  none of these

Ans:  C
Difficulty:  E
Section:  2-3




4.  The average speed of a moving object during a given interval of time is always:
A)  the magnitude of its average velocity over the interval
B)  the distance covered during the time interval divided by the time interval
C)  one-half its speed at the end of the interval
D)  its acceleration multiplied by the time interval
E)  one-half its acceleration multiplied by the time interval.

Ans:  B
Difficulty:  E
Section:  2-4





5.  Two automobiles are 150 kilometers apart and traveling toward each other. One automobile is moving at 60 km/h and the other is moving at 40 km/h. In how many hours will they meet?
A)  2.5
B)  2.0
C)  1.75
D)  1.5
E)  1.25

Ans:  D
Difficulty:  E
Section:  2-4





6.  A car travels 40 kilometers at an average speed of 80 km/h and then travels 40 kilometers at an average speed of 40 km/h. The average speed of the car for this 80 km trip is:
A)  40 km/h
B)  45 km/h
C)  48 km/h
D)  53 km/h
E)  80 km/h

Ans:  D
Difficulty:  M
Section:  2-4





7.  A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A)  0
B)  50 km/hr
C)  100 km/hr
D)  200 km/hr
E)  cannot be calculated without knowing the acceleration

Ans:  A
Difficulty:  E
Section:  2-4





8.  A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The average speed of the car for this round trip is:
A)  0
B)  50 km/h
C)  100 km/h
D)  200 km/h
E)  cannot be calculated without knowing the acceleration

Ans:  B
Difficulty:  E
Section:  2-4





9.  The coordinate of an object is given as a function of time by x = 7t – 3t2, where x is in meters and t is in seconds.  Its average velocity over the interval from t = 0 to t = 2 s is:
A)  5 m/s
B)  –5 m/s
C)  11 m/s
D)  –11 m/s
E)  –14.5 m/s

Ans:  B
Difficulty:  M
Section:  2-4




10.  The coordinate of a particle in meters is given by x(t) = 16t – 3.0t3, where the time t is in seconds. The particle is momentarily at rest at t =
A)  0.75 s
B)  1.3 s
C)  5.3 s
D)  7.3 s
E)  9.3 s

Ans:  B
Difficulty:  M
Section:  2-5

  
Chapter 6


1. The greatest deceleration (of magnitude a) is provided by the maximum friction force (Eq. 6-1, with FN = mg in this case).  Using Newton’s second law, we find

a = fs,max /m = msg.

Eq. 2-16 then gives the shortest distance to stop: |Dx| = v2/2a = 36 m.  In this calculation, it is important to first convert v to 13 m/s.

2. Applying Newton’s second law to the horizontal motion, we have F - mk m g = ma, where we have used Eq. 6-2, assuming that FN = mg (which is equivalent to assuming that the vertical force from the broom is negligible). Eq. 2-16 relates the distance traveled and the final speed to the acceleration: v2 = 2aDx.  This gives a = 1.4 m/s2. Returning to the force equation, we find (with F = 25 N and m = 3.5 kg) that mk = 0.58.

3. The free-body diagram for the bureau is shown to the right. We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion problem (with the person’s push  in the +x direction). Applying Newton’s second law to the x and y axes, we obtain

respectively. The second equation yields the normal force FN = mg, whereupon the maximum static friction is found to be (from Eq. 6-1) .
Thus, the first equation becomes

where we have set a = 0 to be consistent with the fact that the static friction is still (just barely) able to prevent the bureau from moving.

(a) With  and m = 45 kg, the equation above leads to F = 198 N.
To bring the bureau into a state of motion, the person should push with any force greater than this value. Rounding to two significant figures, we can therefore say the minimum required push is F = 2.0 ´ 102 N.

(b) Replacing m = 45 kg with m = 28 kg, the reasoning above leads to roughly .

Note: The values found above represent the minimum force required to overcome the friction. Applying a force greater than  results in a net force in the +x-direction, and hence, non-zero acceleration.

4. We first analyze the forces on the pig of mass m. The incline angle is q.


The +x direction is “downhill.’’ Application of Newton’s second law to the x- and y-axes leads to

Solving these along with Eq. 6-2 (fk = mkFN) produces the following result for the pig’s downhill acceleration:

To compute the time to slide from rest through a downhill distance , we use Eq. 2-15:


We denote the frictionless (mk = 0) case with a prime and set up a ratio:


which leads us to conclude that if t/t' = 2 then a' = 4a. Putting in what we found out above about the accelerations, we have


Using q = 35°, we obtain mk = 0.53.

5. In addition to the forces already shown in Fig. 6-17, a free-body diagram would include an upward normal force  exerted by the floor on the block, a downward  representing the gravitational pull exerted by Earth, and an assumed-leftward  for the kinetic or static friction. We choose +x rightwards and +y upwards. We apply Newton’s second law to these axes:

where F = 6.0 N and m = 2.5 kg is the mass of the block.

(a) In this case, P = 8.0 N leads to

FN = (2.5 kg)(9.8 m/s2) – 8.0 N = 16.5 N.

Using Eq. 6-1, this implies , which is larger than the 6.0 N rightward force – so the block (which was initially at rest) does not move. Putting a = 0 into the first of our equations above yields a static friction force of  f = P = 6.0 N.

(b) In this case, P = 10 N, the normal force is

FN = (2.5 kg)(9.8 m/s2) – 10 N = 14.5 N.

Using Eq. 6-1, this implies , which is less than the 6.0 N rightward force – so the block does move. Hence, we are dealing not with static but with kinetic friction, which Eq. 6-2 reveals to be .

(c) In this last case, P = 12 N leads to FN = 12.5 N and thus to , which (as expected) is less than the 6.0 N rightward force – so the block moves. The kinetic friction force, then, is .

6. The free-body diagram for the player is shown to the right.  is the normal force of the ground on the player,  is the force of gravity, and  is the force of friction. The force of friction is related to the normal force by f = mkFN. We use Newton’s second law applied to the vertical axis to find the normal force. The vertical component of the acceleration is zero, so we obtain FNmg = 0; thus, FN = mg. Consequently,


7. The free-body diagram for the crate is shown to the right. We denote  as the horizontal force of the person exerted on the crate (in the +x direction),  is the force of kinetic friction (in the –x direction),  is the vertical normal force exerted by the floor (in the +y direction), and  is the force of gravity. The magnitude of the force of friction is given by (Eq. 6-2):
fk = mkFN .
Applying Newton’s second law to the x and y axes, we obtain

respectively.

(a) The second equation above yields the normal force FN = mg, so that the friction is

(b) The first equation becomes

which (with F = 220 N) we solve to find


Note: For the crate to accelerate, the condition  must be met. As can be seen from the equation above, the greater the value of , the smaller the acceleration with the same applied force.

8. To maintain the stone’s motion, a horizontal force (in the +x direction) is needed that cancels the retarding effect due to kinetic friction. Applying Newton’s second to the x and y axes, we obtain

respectively. The second equation yields the normal force FN = mg, so that (using Eq. 6-2) the kinetic friction becomes fk = mk mg. Thus, the first equation becomes


where we have set a = 0 to be consistent with the idea that the horizontal velocity of the stone should remain constant. With m = 20 kg and mk = 0.80, we find F = 1.6 ´ 102 N.

9. We choose +x horizontally rightwards and +y upwards and observe that the 15 N force has components Fx = F cos q and Fy = – F sin q.

(a) We apply Newton’s second law to the y axis:


With mk = 0.25, Eq. 6-2 leads to fk = 11 N.

(b) We apply Newton’s second law to the x axis:

.

Since the result is positive-valued, then the block is accelerating in the +x (rightward) direction.

10. (a) The free-body diagram for the block is shown below, with being the force applied to the block,  the normal force of the floor on the block,  the force of gravity, and  the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up. The equations for the x and the y components of the force according to Newton’s second law are:
Now f =mkFN, and the second equation gives FN = mg Fsinq, which yields . This expression is substituted for f in the first equation to obtain

F cos qmk (mgF sin q) = ma,
so the acceleration is
.

(a) If  and  then the magnitude of  has a maximum value of

                        

On the other hand,  Therefore,  and the block remains stationary with .

(b) If  and  then the magnitude of  has a maximum value of

                        

In this case,  Therefore, the acceleration of the block is
         

11. (a) The free-body diagram for the crate is shown on the right.  is the tension force of the rope on the crate,  is the normal force of the floor on the crate,  is the force of gravity, and  is the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up. We assume the crate is motionless. The equations for the x and the y components of the force according to Newton’s second law are:

               T cos qf = 0
                                  

where q = 15° is the angle between the rope and the horizontal. The first equation gives f = T cos q and the second gives FN = mgT sin q. If the crate is to remain at rest, f must be less than ms FN, or T cos q < ms (mgT sinq). When the tension force is sufficient to just start the crate moving, we must have

T cos q = ms (mgT sin q).

We solve for the tension:





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